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Hi there
I need help with the restrictions on the variables of this question:

Simplify. State any restrictions on the variables.


So my answer is: log(x+4/x-3) which is correct.

Now for the restrictions, I have:
x<-4 and x>3

However the back of the book says the answer is only x>3 and not x<-4 for the restrictions of the variables. But when I sub in any number such as -5 into the ORIGINAL log equation, i get a defined and positive log of x value. So? is mine correct or theirs? Explain fully.

  • Math - ,

    Sorry the question is:


  • Math - ,

    first of all
    log(x^2+7x+12)/log(x^2-9) is NOT equal to
    log((x+4)/(x-3)) like you say
    If the back of your book has that answer, then they are definitely WRONG

    log [(x^2+7x+12)/(x^2-9)]
    would be equal to log((x+4)/(x-3))

    to find the restrictions you have to consider the numerator and denominator
    remember we can only take logs of positive numbers
    at the top:
    log(x^2+7x+12) = log[(x+3)(x+4)]
    x < -4 OR x > -3

    at the bottom:
    log(x^2 - 9) = log[(x-3)(x+3)]
    x < -3 or x > 3

    but both of these conditions must be met, so

    x < -4 OR x > -3 AND x < -3 or x > 3

    which leaves you with x < -4 or x > 3

    notice that between -4 and +3 we get a negative for either one or the other function

    another way to understand the restriction would be:
    graph both y = (x+3)(x+4) and
    y = (x+3)(x-3)
    any x where either or both of those graphs dip below the x-axis would not be allowed.

    so yes, they are wrong with their restrictions.

  • Math - ,

    ARGHHHH! saw your correction only after I posted.

    So my answer would match your first posting, and it would clearly be a more interesting question.

  • Math - ,

    so now
    log [(x+4)(x+3)/((x-3)(x+3))]
    = log(x+4) + log(x+3) - log(x+3) - log(x-3)

    now each of those terms must be defined.
    so x>-4 AND x>-3 AND x>3

    which is x > 3

    notice that we canceled out log(x+3)
    but for all x ≤ -3 that term would be undefined. So can we really do algebra with undefined numbers??

    I know this is a bit tricky, since -5 (or any x < -4) would give a positive result in the original algebraic expression, but it would not work in the individuals terms.

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